power supplies

ieee paper on types of power supplies in emi
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   Application Note 140October 2013Basic Concepts of Linear Regulator and Switcing!ode ower Supplies #enr$ %& 'ang ABSTRACT (is article e)plains te basic concepts of linear regulatorsand switcing *ode power supplies +S! S,& -t is ai*ed ats$ste* engineers wo *a$ not be .er$ fa*iliar wit power suppl$ designs and selection& (e basic operatingprinciples of linear regulators and S! S are e)plained andte ad.antages and disad.antages of eac solution arediscussed& (e buc/ stepdown con.erter is used as ane)a*ple to furter e)plain te design considerations of aswitcing regulator& INTRODUCTION (oda$s designs reuire an increasing nu*ber of power railsand suppl$ solutions in electronics s$ste*s wit loadsranging fro* a few *A for standb$ supplies to 100A for  AS-C .oltage regulators& -t is i*portant to coose teappropriate solution for te targeted application and to *eetspecified perfor*ance reuire*ents suc as ig efficienc$tigt printed circuit board + CB, space accurate outputregulation fast transient response low solution cost etc& ower *anage*ent design is beco*ing a *ore freuent andcallenging tas/ for s$ste* designers *an$ of wo* *a$not a.e strong power bac/grounds&  A power con.erter generates output .oltage and currentfor te load fro* a gi.en input power source& -t needs to*eet te load .oltage or current regulation reuire*entduring stead$state and transient conditions& -t also *ustprotect te load and s$ste* in case of a co*ponentfailure& epending on te specific application a designer can coose eiter a linear regulator +LR, or a switcing*ode power suppl$ +S! S, solution& (o *a/e te bestcoice of a solution it is essential for designers to befa*iliar wit te *erits drawbac/s and design concerns of eac approac& (is article focuses onnonisolated power suppl$ applicationsand pro.ides anintroduction to teir operation and designbasics& LINEAR REGULATORSHow a Linear Regulator Works Lets start wit a si*plee)a*ple& -n an e*beddeds$ste* a 125 bus rail isa.ailable fro* te frontendpower suppl$& On tes$ste* board a 3&35.oltage is needed to power an operational a*plifier +opa*p,& (e si*plestapproac to generate te3&35 is to use a resistor di.ider fro* te 125 bus assown in 6igure 1& oes itwor/ well7 (e answer isusuall$ no& (e op a*ps5 CC  pin current *a$ .ar$under different operatingconditions& -f a fi)ed resistor di.ider is used te -C 5 CC .oltage .aries wit load&Besides te 125 bus input*a$ not be well regulated&(ere *a$ be *an$ oter loads in te sa*e s$ste*saring te 125 rail&Because of te busi*pedance te 125 bus.oltage .aries wit te busloading conditions& As aresult a resistor  125C B8SR13&35 5 CC R259 –  AN140 601 Figure ! Resistor Di i#er Generates $!$% DC  &ro' (% Bus In)ut L  L( L(C L(! Linear (ecnolog$ L(spice :!odule ol$ ase and te Linear logo are   registered trade*ar/s and L(powerCA is a trade*ar/ of Linear (ecnolog$ Corporation& All oter trade*ar/s are te propert$ of teir respecti.e owners& an140fa AN140-1   Application Note 140 di.ider cannotpro.ide aregulated3&35 tote opa*p toensure itsproper operation&(ereforeadedicated.oltageregulationloop isneeded& As sownin 6igure2 tefeedbac/loopneeds toad;ust tetopresistor R1 .aluetod$na*icall$ regulatete 3&35on 5 CC & 1255 –  AN140 602 Figure (!Fee#*a+k Loo) A#,usts Series Resistor R %alue to Regulate $!$% (is /ind of .ariableresistor canbei*ple*entedwit a linear regulator assown in6igure 3& Alinear regulator operates abipolar or field effectpower transistor +6<(, in itslinear *ode&So tetransistor wor/s as a.ariableresistor inseries witte outputload& (oestablis tefeedbac/loopconceptuall$an error a*plifier senses teC output.oltage .ia asa*plingresistor networ/ R  A and R B  tenco*pares tefeedbac/.oltage 5 6B wit areference.oltage 5 R<6 &(e error a*plifier output.oltage dri.este base of te seriespower transistor .iaa currenta*plifier&=en eiter te input5 B8S  .oltagedecreases or te loadcurrentincreases te5 CC  output.oltage goesdown& (efeedbac/.oltage 5 6B decreases aswell& As aresult tefeedbac/error a*plifier and currenta*plifier generate*ore currentinto te baseof tetransistor >1&(is reduceste .oltagedrop 5 C< and encebrings bac/te 5 CC output.oltage sotat 5 6B euals 5 R<6 &On te oter and if te5 CC  output.oltage goesup in asi*ilar wa$te negati.efeedbac/circuitincreases5 C<  L-N<AR R<?8LA(OR C8RR<N( A! L-6-<R <RROR  A! L-6-<R Figur e $! A Linear Regulator I')le'entsa %aria*le Resistor to Regulate Out)ut %oltage to ensureteaccurateregulationof te3&35output& -nsu**ar$an$.ariationof 5 O  isabsorbedb$ telinear regulator transistors 5 C< .oltage& Sote output.oltage 5 CC is alwa$sconstant andwellregulated& W-. Use Linear Regulator s/ (e linear regulator asbeen widel$used b$industr$ for$ longti*e& -t waste basis for te power suppl$industr$ untilswitcing*ode power suppliesbeca*epre.alentafter te1@0s& <.entoda$ linear regulatorsare stillwidel$ usedin a widerange of applications& -n addition toteir si*plicit$ of use linear regulatorsa.e oter perfor*ancead.antages& ower *anage*entsuppliersa.ede.eloped*an$integratedlinear regulators& At$picalintegratedlinear regulator needs onl$5 -N  5 O8( 6B andoptional ?Npins& 6igure 4sows at$pical 3pinlinear regulator teL(103wic wasde.eloped*ore tan 20$ears ago b$Linear (ecnolog$& -tonl$ needsan inputcapacitoroutputcapacitor andtwo feedbac/resistors toset te output.oltage& Al*ost an$electricalengineer candesign asuppl$ wittese si*plelinear regulators& an140fa AN140-2   Application Note 140 5-N  &D5 ER<>8-R< 6OR S(AB-L-(F Figur e 0! Integr ate# Linear  Regulator  E1a')le2 3!4A Linear Regulator wit-Onl. T-r ee5ins One Draw*a+k 6 A Linear Regulator Can Burn a Lot o& 5ower   A *a;or drawbac/ of using linear regulatorscan be tee)cessi.epower dissipation of its seriestransistor >1operating ina linear *ode& Ase)plainedpre.iousl$ alinear regulator transistor isconceptuall$a .ariableresistor&Since all teload current*ust passtroug teseriestransistor itspower dissipation is Loss  G +5 -N  H 5 O , - O & -ntis case teefficienc$ of a linear regulator canbe uic/l$esti*ated b$I η  G O8( 8( LR J O8( 8( LOSS So in te6igure 1e)a*plewen teinput is 125and output is3&35 telinear regulator efficienc$ is ;ust 2K&D& -ntis caseK2&D of teinput power is ;ust wastedandgenerateseat in teregulator&(is *eanstat tetransistor *ust a.e teter*alcapabilit$ toandle itspowerM eatdissipation atworst case at*a)i*u* 5 -N and full load&So te sie of te linear regulator andits eat sin/*a$ be large
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